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If α is an r-cycle, show that α ^r = (1)

If α is an r-cycle, show that α ^r = (1)

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Answer #1

Actually this statement tell us the length of a cycle is the order of the cycle.

Now we prove it.

Now suppose the r - cycle, a= (b0, b1, b2, b3....... br). It's order can't be less than n because if 0<k<n then a^k(b0) =bk and it's implicit in the concept of a cycle that b0 is not equal to bk. On the other hand, the a^r is surely equal to identity permutations because it acts on each element of the cycle by moving it around the entire cycle one time. Again we know that in permutations group, a^k(bj) =b(j+k)mod n and also since n=0(modn) then we have (j+n) modn=j, whenever j<k<n and j=0. And all elements that are not in the cycle at all are left along by a and therefore a^r. Since, a^k is not equal to identity when 0<k<n, hence a has order r. Hence a^r=1.(proved).

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